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Math Help - [SOLVED] Integral formulas

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Integral formulas

    Hi MHF!
    I need your help for some questions.
    First, what is the method to find out that \int \frac{dx}{1+x^2}=arctan(x)+C ? I find it stupid to learn it by heart. There are plenty of other formulas to know, so I would like to know how to recover them without learning all of them.
    Second, I have to find \int \frac{dx}{\sqrt{x^2-9}} in term of trigonometric or hyperbolic functions. Looking at my book, I found out that it is equal to ln|x+\sqrt{x^2-9}|+C. Which recalls me of an inverse of cosh^{-1} but cannot find exactly to what it's equal, so I'd appreciate help for this one.
    And finally \int \frac{dx}{\sqrt{4-x^2}}=arcsin\big(\frac{x}{2}\big)+C according to my book. But once again, how can I know it's worth an inverse of the sine function? I hope I don't have to learn this by heart.
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  2. #2
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    Here is a way to look at it and probably the method commonly used.

    \frac{d}{dx}[tan^{-1}(x)]=\frac{1}{1+x^{2}}

    y=tan^{-1}(x), \;\ x=tan(y)

    Differentiating the right one implicitly gives us:

    \frac{d}{dx}[x]=\frac{d}{dx}[tan(y)]

    1=sec^{2}(y)\cdot\frac{dy}{dx}

    \frac{dy}{dx}=\frac{1}{sec^{2}(y)}=\frac{1}{sec^{2  }(tan^{1}(x))}.....[1]

    Using the triangle we see that:

    sec^{2}(tan^{-1}(x))=1+x^{2}

    And [1] simplifies to \frac{dy}{dx}=\frac{1}{1+x^{2}}

    Therefore, \int\frac{1}{1+x^{2}}=tan^{-1}(x)

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    To find \int\frac{1}{\sqrt{x^{2}-9}}dx

    let x=3sec{\theta}, \;\ dx=3sec{\theta}tan{\theta}d{\theta}

    When you make the subs and simplify it whittles down to

    \int sec{\theta}d{\theta}

    Then, to get back in terms of x just resub {\theta}=sec^{-1}(\frac{x}{3})

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~

    \int\frac{1}{\sqrt{4-x^{2}}}dx

    This one simplifies wonderfully. Let x=2sin{\theta}, \;\ dx=2cos{\theta}d{\theta}

    Make the subs and you finally get:

    \int d{\theta}={\theta}

    What is theta?. {\theta}=sin^{-1}(x/2)

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~

    To tell when to use what, just remember the formulas:

    \sqrt{a^{2}-x^{2}}, then use x=a\cdot sin{\theta}

    \sqrt{a^{2}+x^{2}}, then use x=a\cdot tan{\theta}

    \sqrt{x^{2}-a^{2}}, then use x=a\cdot sec{\theta}

    Can you see whey we use those when we do?. Because they result in the proper identities that get rid of the radical.
    Last edited by galactus; November 24th 2008 at 06:38 AM.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks a lot, I'll be analyzing this closely.
    You made an error of typo in the line " \frac{dy}{dx}=\frac{1}{sec^{2}(y)}=\frac{1}{sec^{2  }(tan^{1}(x))}.....[1]" Must be tan^{-1}. Otherwise I think all is O.K. Thanks so much!!!
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Smile

    Quote Originally Posted by arbolis View Post
    And finally \int \frac{dx}{\sqrt{4-x^2}}=arcsin\big(\frac{x}{2}\big)+C according to my book. But once again, how can I know it's worth an inverse of the sine function? I hope I don't have to learn this by heart.
    Let y = \arcsin x. Then we have

    x = \sin y

    \Rightarrow \cos y\frac{dy}{dx} = 1

    \Rightarrow \frac{dy}{dx} = \frac1{\cos y} = \frac1{\sqrt{1 - \sin^2 y}} = \frac1{\sqrt{1 - x^2}}

    The other inverse functions are similar.
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  5. #5
    Senior Member nikhil's Avatar
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    Basic thinking

    hi!arbolis,hope ur having a great time.
    Hmmmmm..... You have asked a very basic question.though i can tell you how to integrate some function by a definite procedure (like for your first integral question put x=tany,for second put X=3secy etc) but i think i should give you basics that are behind the procedures
    The main problem is that unlike differentiation there is no definite first principle for integration.when we perform integration,we take it as a reverse process of differentiation. Like
    d(sinx)/dx=cosx
    therefor
    integral cosxdx=sinx.
    So as you can see if you know the derivative of a function then the integral of its derivative is the function itself.so whenever we need to integrate a complex looking function we try to make it less complex by substitution.
    Like A= \int \frac{dx}{1+x^2}
    let x=tany
    differentiating both sides we get
    dx=(secy)^2
    now substituting these value we get
    A= <br />
\int\frac{(secy)^{2} dy}{1+(tany)^2}<br />
    but 1+(tany)^2=(secy)^2
    so A= <br />
\int dy<br />
[so because of the substitution integral became easy]
    so A=y+c
    since x=tany
    y=arctanx
    so A=arctanx+c
    i hope this will help in widening your thinking.
    REMEMBER it may not be possible to solve all questions by substitution.so like differentiation there are some other procedures[like partial fraction method}.
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  6. #6
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    It is really helpful to memorize some basic forms. For example the derivative of \sin^{-1} (on (-1,1)). The way we do this is to noting that \sin (\sin^{-1} x ) = x thus (\sin^{-1} x) \cos (\sin^{-1} x) = 1 by the chain rule. But, \cos (\sin^{-1} x) = \sqrt{1-x^2} which means (\sin^{-1} x) ' = \tfrac{1}{\sqrt{1-x^2}}. Here is an example, \smallint \tfrac{dx}{\sqrt{9-x^2}}. We can factor this, \tfrac{1}{3}\smallint \tfrac{dx}{\sqrt{1-(\frac{x}{3})^2}}. Now let t=\tfrac{x}{3} and we get \smallint \tfrac{dt}{\sqrt{1-t^2}} = \sin^{-1} t + k = \sin^{-1} \tfrac{x}{3}+k.

    Now find a similar form for \sinh^{-1} x. Again, \sinh (\sinh^{-1} x) = x by the chain rule we get (\sinh^{-1} x)' \cosh(\sinh^{-1} x) = 1. Now note that \cosh(\sinh^{-1} x) = \sqrt{x^2+1}*. For example, \smallint \tfrac{dx}{\sqrt{x^2+4}} can be solved in a similar way as above problem. Inside of doing all those substitutions.


    *)Let f(x) = \cosh(\sinh^{-1} x) and see that f(x)>0. But then f^2(x) = \cosh^2 (\sinh^{-1} x) = 1 + \sinh^2(\sinh^{-1} x) = 1+x^2. Thus, f(x) = \sqrt{x^2+1}.
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